Find the magnitude and direction of the velocity

[inigooo]

Homework Statement

A ball is hanging from a cord of negligible weight. A secondary ball with the same weight as the first ball is measured 3m above the hanging ball, and just touching the cord, is released from rest and gains its velocity before striking the hanging ball. If the effect of friction is negligible, and the impact is perfectly elastic, what is
a) the magnitude of the velocity of the secondary ball in terms of its velocity before contact;
b) the direction of the velocity of the secondary ball in terms of its velocity before contact

Relevant Equations

My answers are a) 7.67 m/s and b) downward, however my teacher says it’s wrong

V^2= u^2 +2gh

 

[malawi_glenn]

Show your work, not just your answers

 

[inigooo]

v^2= u^2 +2gh
v^2= 0^2+2(9.81)(3)
v=sqrt(2x9.81x3)
v=sqrt(58.86)
v=7.67m/s, downward
here is my solution

 

[malawi_glenn]

Given one signigicant digit, don't answer with three.

Are you asked about the magnitude and direction of the velocity directly after the collision with the primary ball?

I do not think you are supposed to answer with numbers

a) the magnitude of the velocity of the secondary ball in terms of its velocity before contact;
b) the direction of the velocity of the secondary ball in terms of its velocity before contact

 

[inigooo]

no, it is ’in terms of its velocity before contact’

 

[malawi_glenn]

no, it is ’in terms of its velocity before contact’

Such question means that you should find a ratio or similar. Let's say the velocity just before the impact was $\vec v_1$ and the velocity just after the impact was $\vec v_2$. You should just answer what $|\vec v_2| / |\vec v_1|$ is.

Similarly for the b)-part. Is the direction of $\vec v_2$ same, opposite, 90 degree, etc, compared to $\vec v_1$?

The problem says that the collision is elastic. Do you know what that means and how to use that information here?

 

[inigooo]

The problem says that the collision is elastic. Do you know what that means and how to use that information here?

 

[inigooo]

I dont

 

[malawi_glenn]

I dont

Then look it up in your textbook.

 

[inigooo]

Then look it up in your textbook.

Pls. How can i get the magnitude and direction of the velocity of the secondary ball in terms of its velocity before contact?

 

[malawi_glenn]

Pls. How can i get the magnitude and direction of the velocity of the secondary ball in terms of its velocity before contact?

State what an elastic collison is.
https://www.physicsforums.com/threads/homework-help-guidelines-for-students-and-helpers.686781/

 

[inigooo]

V^2=u^2 +2gh
v= sqrt(2x9.81x3)
v=7.67m/s
a) 7.67m/s
cos theta = 1
or theta = 90 degree
b) along 90 degrees with vertical line

however, this computation is for after collision. how can i compute for the magnitude and direction of the velocity of the secondary ball in terms of its velocity before contact?

 

[inigooo]

State what an elastic collison is.
https://www.physicsforums.com/threads/homework-help-guidelines-for-students-and-helpers.68678

Where bodies separates completely after collision. Linear momentum is conserved and total kinetic energy IS conserved

 

[malawi_glenn]

Where bodies separates completely after collision. Linear momentum is conserved and total kinetic energy IS conserved

Yes. How can you use these two conditions on this problem?

 

[inigooo]

Yes. How can you use these two conditions on this problem?

The total kinetic energy before and after collision will remain the same, and the total momentum along a particular direction before and after collision will remain the same

 

[malawi_glenn]

The total kinetic energy before and after collision will remain the same, and the total momentum along a particular direction before and after collision will remain the same

Write down the equations for the linear momentum before and after the collision, same with kinetic energy. Then solve that system of equations.
Note: what will happen to the string supporting the primary ball?

 

[inigooo]

Write down the equations for the linear momentum before and after the collision, same with kinetic energy. Then solve that system of equations.
Note: what will happen to the string supporting the primary ball?

1/2m1v1^2+1/2m2v2^2=1/2m1v1’^2+1/2m2v2’^2
m1vx1+1/2m2vx2=m1v’x1 + m2v’y2
m1vy1+1/2m2vy2=m1v’y1 + m2v’y2
v2^2=u^2+2gh
v2=sqrt(6x9.81)
0+1/2mv2^2=1/2mv’1^2+1/2mv’2^2
v2^2=v’1^2 + v’2^2
0+0=mv’1-mv’2sintheta
sintheta=v’1/v’2
mv2=mv’2costheta
costheta= v2/v’2
v2^2-v’2^2+v2^2=v’2^2
v’2=v2
v’2=sqrt(58.86)
costheta=1 or theta = 90Degrees

 

[malawi_glenn]

Can you use LaTeX?

What does a prime quantity refer to?

 

[inigooo]

Can you use LaTeX?

What does a prime quantity refer to?

After collision

 

[malawi_glenn]

After collision

https://www.physicsforums.com/help/latexhelp/

 

[inigooo]

https://www.physicsforums.com/help/latexhelp/

1/2 m_1v_1^2 + 1/2 m_2v_2^2 = 1/2 m_1v_1’^2 + 1/2 m_2v_2’^2
m_1v_x1 + 1/2 m_2v_x2 = m_1v’_x1 + m_2v’_y2
m_1v_y1 + 1/2 m_2v_y2 = m_1v’_y1 + m_2v’_y2
v_2^2 = u^2 +2gh
v_2=sqrt(6x9.81)
0+1/2 mv_2^2 = 1/2 mv’_1^2 + 1/2 mv’_2^2
v_2^2 = v’_1^2 + v’_2^2
0+0= mv’_1 - mv’_2 sintheta
sintheta = v’_1 / v’_2
mv_2=mv’_2 costheta
costheta= v_2 / v’_2
v_2^2 - v’_2^2 + v_2^2 = v’_2^2
v’_2 = v_2
v’_2 = sqrt(58.86)
costheta=1 or theta = 90Degrees

 

[inigooo]

I don't know what the final answer is

1/2 m_1v_1^2 + 1/2 m_2v_2^2 = 1/2 m_1v_1’^2 + 1/2 m_2v_2’^2
m_1v_x1 + 1/2 m_2v_x2 = m_1v’_x1 + m_2v’_y2
m_1v_y1 + 1/2 m_2v_y2 = m_1v’_y1 + m_2v’_y2
v_2^2 = u^2 +2gh
v_2=sqrt(6x9.81)
0+1/2 mv_2^2 = 1/2 mv’_1^2 + 1/2 mv’_2^2
v_2^2 = v’_1^2 + v’_2^2
0+0= mv’_1 - mv’_2 sintheta
sintheta = v’_1 / v’_2
mv_2=mv’_2 costheta
costheta= v_2 / v’_2
v_2^2 - v’_2^2 + v_2^2 = v’_2^2
v’_2 = v_2
v’_2 = sqrt(58.86)
costheta=1 or theta = 90Degrees

i don't know what the final answer is

 

[malawi_glenn]

1/2 m_1v_1^2 + 1/2 m_2v_2^2 = 1/2 m_1v_1’^2 + 1/2 m_2v_2’^2
m_1v_x1 + 1/2 m_2v_x2 = m_1v’_x1 + m_2v’_y2
m_1v_y1 + 1/2 m_2v_y2 = m_1v’_y1 + m_2v’_y2
v_2^2 = u^2 +2gh
v_2=sqrt(6x9.81)
0+1/2 mv_2^2 = 1/2 mv’_1^2 + 1/2 mv’_2^2
v_2^2 = v’_1^2 + v’_2^2
0+0= mv’_1 - mv’_2 sintheta
sintheta = v’_1 / v’_2
mv_2=mv’_2 costheta
costheta= v_2 / v’_2
v_2^2 - v’_2^2 + v_2^2 = v’_2^2
v’_2 = v_2
v’_2 = sqrt(58.86)
costheta=1 or theta = 90Degrees

Almost. You are missing some important code to make it appear as latex

 

[inigooo]

Almost. You are missing some important code to make it appear as latex

Is the answer a) unchanged, b) upward?

 

[malawi_glenn]

Is the answer a) unchanged, b) upward?

Why do you think that is the correct answer?

 

[inigooo]

Why do you think that is the correct answer?

Since it says that ‘in terms of its velocity before contact’? Then the magnitude of its velocity is 0m/s or unchanged, then the direction is upward or opposite to v1?

 

[malawi_glenn]

Since it says that ‘in terms of its velocity before contact’? Then the magnitude of its velocity is 0m/s or unchanged, then the direction is upward or opposite to v1?

Cant it be 40% of initial speed?

 

[inigooo]

Cant it be 40% of initial speed?

40%?

 

[malawi_glenn]

40%?

Or 75%

 

[inigooo]

Or 75
why would the the magnitude of its velocity is 75% or 40% of initial speed?

 

[inigooo]

Since it says that ‘in terms of its velocity before contact’? Then the magnitude of its velocity is 0m/s or unchanged, then the direction is upward or opposite to v1?

Is this wrong?

 

[malawi_glenn]

Is this wrong?

Why do you think your answer is correct?

 

[inigooo]

Why do you think your answer is correct?

Since it says that ‘in terms of its velocity before contact’. Then the magnitude of its velocity is 0m/s or unchanged, then the direction is upward or opposite to v1.

 

[inigooo]

Since it says that ‘in terms of its velocity before contact’. Then the magnitude of its velocity is 0m/s or unchanged, then the direction is upward or opposite to v1.

Pls. Help me point out what is wrong, if it is wrong…

 

[malawi_glenn]

Then the magnitude of its velocity is 0m/s or unchanged, then the direction is upward or opposite to v1.

Why is it so?

If it is 0m/s how can ut have a direction?

 

[inigooo]

Why is it so?

If it is 0m/s how can ut have a direction?

Ohhh. so it should be, no direction? Then a) 0m/s and b) no direction?

 

[malawi_glenn]

Ohhh. so it should be, no direction? Then a) 0m/s and b) no direction?

Why should it be 0m/s?

Look i know you only want the correct answer and move on. But that is not how we operate on this forum

 

[inigooo]

Why should it be 0m/s?

Because (v’2)/(v’1)
(7.67m/s)/(0m/s)= 0m/s?
if it’s wrong, what should i do to get the magnitude of the velocity in terms of its velocity before contact?

 

[malawi_glenn]

Because (v’2)/(v’1)
(7.67m/s)/(0m/s)= 0m/s?
if it’s wrong, what should i do to get the magnitude of the velocity in terms of its velocity before contact?

Divide by 0?
How do you know that the secondary ball has velocity 0m/s after collision? Can you calculate it?

 

[inigooo]

Divide by 0?
How do you know that the secondary ball has velocity 0m/s after collision? Can you calculate it?

how can i get the v’1?

 

[inigooo]

how can i get the v’1?

pls. what should i do to get the final answers

 

[Steve4Physics]

@inigooo, Can I chip in. The question says:

“A secondary ball with the same weight as the first ball is ... just touching the cord” [my underlining]​

The wording in the question is (IMO) poor. I guess it should also say that the two balls have the same radius. So the centre of the upper ball is above the edge of the lower ball.

A good diagram is needed showing the positions of the balls at the moment of impact.

I suggest stepping-back a bit; revise/read-up how to solve simple ‘oblique collision’ problems. There is plenty of help available, e.g. a YouTube search for “elastic oblique collision in two dimensions” gives a number of videos. The first video I got (which is not necessarily the best) was:


EDIT: Can I add this, on edit. I'm not saying that the impact is a simple oblique collision because the presence of the (presumably inextensible) chord is a major constraint. However, considering the line-of-centres and the tangent plane between the balls - as we would for a simple oblique collision - is important.

 

[malawi_glenn]

Something like this:

Because if the balls have to be treated as pointmasses, one have to assume something regarding the chord.

 

[kuruman]

Something like this:
View attachment 305409

Because if the balls have to be treated as pointmasses, one have to assume something regarding the chord.

Why point masses?

Here is the diagram at the time of the collision assuming that the balls have the same radius $R$. It seems to me that the momentum of the two-ball system is not conserved in the vertical direction because the string constrains part of the system from moving in that direction. Assuming that there is no friction while the balls are in contact, the top ball delivers an impulse directed along the center-to-center direction which can be calculated from the geometry.

 

[malawi_glenn]

Why point masses?

I mean something like this

so that it effectively becomes a 1-d problem. But then one has to assume something about the chord.

 

[kuruman]

If this were a 1-d problem, then why ask for the direction of the velocity in part (b)?

 

[haruspex]

Why point masses?

Here is the diagram at the time of the collision assuming that the balls have the same radius $R$. It seems to me that the momentum of the two-ball system is not conserved in the vertical direction because the string constrains part of the system from moving in that direction. Assuming that there is no friction while the balls are in contact, the top ball delivers an impulse directed along the center-to-center direction which can be calculated from the geometry.

View attachment 305408

I'm not sure the problem is fully defined. There are in effect two impacts, one between the balls, which we are told is elastic, but one also in the sudden tensioning of the string. We can assume the string is inextensible but is it elastic or fully inelastic?

Edit: in the case of an elastic string, we would also need to know the ratio between the two "spring constants". So assume fully inelastic.

Edit 2:
Correction: the solution @kuruman and I got independently assumes no loss of KE, so everything is assumed to be perfectly elastic.
Our other assumption is that the struck ball only moves sideways. Whether e.g. the string's spring constant being much greater than that of the balls would make that true I am not sure. More generally, it seems feasible that the struck ball would also acquire some vertically upward velocity.

But what if we take the string to be perfectly inelastic? Still puzzling over that one.

 

[malawi_glenn]

If this were a 1-d problem, then why ask for the direction of the velocity in part (b)?

Same/oppsite are directions no?

 

[inigooo]

I'm not sure the problem is fully defined. There are in effect two impacts, one between the balls, which we are told is elastic, but one also in the sudden tensioning of the string. We can assume the string is inextensible but is it elastic or fully inelastic?

Edit: in the case of an elastic string, we would also need to know the ratio between the two "spring constants". So assume fully inelastic.

Hello. Is the problem unsolvable?

 

[inigooo]

Same/oppsite are directions no?

But the direction has to be in degrees with respect to North,south, east, or west